Re: Curvature by Cartan's method?

These

nicola_ambrosetti@xxxxxxxx wrote:
(1) do^a + w^a_b /\ o^b = 0
R^a_b = dw^a_b + w^a_c /\ w^c_b = R^a_bcd o^c /\ o^d

are generic to any frame and don't require orthonormal frames at all.

In general, you have
w^a_b = w^a_{cb} o^c
with the connection coefficients expressed in terms of the metric h and
structure coefficients by
w^a_{cb} = g^{ad} w_{cb,d}
w_{cb,d} = 1/2 (e_c g_{bd} + e_b g_{cd} - e_d g_{bc})
+ 1/2 (f_{cb,d} + f_{dc,b} + f_{db,c})
with the structure coefficients given by
f_{ab,c} = g_{cd} f^d_{ab}
and determined through the commutators
[e_a,e_b] = f^c_{ab} e_c.

For orthonormal frames, g_{ab} = g(e_a,e_b) and g^{ab} = g(o^a,o^b) are
constant, so that the connection coefficients are found directly from
the structure coefficients and, ultimately, directly from the
commutators.

You can always try to work backwards to solve for w^a_b in (1), using
the anti-symmetry rule g_{ac} w^c_b = -g_{bc} w^c_a, but it's more
direct just to write out the commutators and find the connection
coefficient from these.

There's actually a relatively simple form for the curvature scalar R in
terms of the various f's and their first derivatives, by the way.
Likewise, the Ricci tensor can be written down directly in terms of the
f's, though it's a bit more complex.

.